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PROBLEM 16.128 (Continued)
Kinematics:
Equating i-terms:
Substitute into (2)(5)(7)→(4):
BA
=vv
( )
B/A
0.2
0.8
AB
AB
ω
π
+×
= ×
= −
ωr
kj
i
( )
C/B
0.8 0.4
0.8 0.4
C B BC
C BC
C BC
v
v
πω
πω
=−×
=−+
j i+ k j
ji
2
BC
ωπ
= −
2/AAB B AB
yy
,
0.2 0.2 5
x BC x
Ca−+
( )
BC
x
C
a
−=
15
0
x
C=
PROBLEM 16.129
The 4-kg uniform slender bar BD is attached to bar AB and a wheel
of negligible mass which rolls on a circular surface. Knowing that at
the instant shown bar AB has an angular velocity of 6 rad/s and no
angular acceleration, determine the reaction at Point D.
PROBLEM 16.129 (Continued)
Kinetics:
4 kgm=
22 2
11
(4)(0.75) 0.1875 kg m
12 12
G
I mL= = = ⋅
( ):
MMΣ=Σ
PROBLEM 16.130
The motion of the uniform slender rod of length L = 0.5 m and mass m = 3 kg is
guided by pins at A and B that slide freely in frictionless slots, circular and
horizontal, cut into a vertical plate as shown. Knowing that at the instant shown
the rod has an angular velocity of 3 rad/s counter-clockwise and θ = 30°,
determine the reactions at Points A and B.
PROBLEM 16.130 (Continued)
Equating the two expressions for
B
a
gives
2
2
eff
2 22
C C Gx Gy
0.5
(3)(9.81) sin30 0.0625
20.5
(0.64952 9.00) cos30
2
0.5
(5.8457 0.375 ) sin30
2
a
a
a
°=
++ °
−− °
2
3.67875 0.25 1.21784
9.8436 rad/s
a
a
= +
=
[(0.64952)(9.8436) 9.00]
G
m= +a
[5.8457 (0.375)(9.8436)]+−
[15.394 N]=
[2.1544 N]
PROBLEM 16.130 (Continued)
:
mAΣ=Σ =FF a
B+
mg+
m=a
PROBLEM 16.131
At the instant shown, the 20 ft long, uniform 100-lb pole ABC has an
angular velocity of 1 rad/s counterclockwise and Point C is sliding to the
right. A 120-lb horizontal force P acts at B. Knowing the coefficient of
kinetic friction between the pole and the ground is 0.3, determine at this
instant (a) the acceleration of the center of gravity, (b) the normal force
between the pole and the ground.
PROBLEM 16.131 (Continued)
y
FΣ=
2
eff
( ) : sin10 cos10
22
y
ml ml
F N mg
aω
Σ − =− °− °
2
2
sin10 cos10
22
100 100
ml ml
N mg
aω
° += − °
PROBLEM 16.132
A driver starts his car with the door on the passenger’s side wide
open
( 0).
θ
=
The 80-lb door has a centroidal radius of gyration
12.5 in.,k=
and its mass center is located at a distance
22 in.r=
from its vertical axis of rotation. Knowing that the
driver maintains a constant acceleration of 6 ft/s2, determine the
angular velocity of the door as it slams shut
( 90 ).
θ
= °
PROBLEM 16.132 (Continued)
( )
( )
( )
22
12
22
12.5 22
12 12
ft cos
ft ft
0.41234 cos
A
A
A
a
d
d
d ad
ω
ωθ
θ
ωω θ θ
=
+
=
/2
00
/2
20
0
2
(0.4124 )cos
10.41234 |sin |
2
0.82468
f
f
A
A
fA
d ad
a
a
ωπ
ω
π
ωω θ θ
ωθ
ω
=
=
=
∫∫
(1)
Given data:
2
6 ft/s
A=a
2
22
0.82468(6)
4.948 rad /s
f
ω
=
=
2.22 rad/s
f
ω
=
